I want to show that for every $z \in [0,1)$ there is a subsequence of $d(n)$ that converges against $z$.

Notes:

I am having serious problems of finding such a subsequence, first I see what the sequence does. If you take for example the subsequence $a(n) = d(10^n)$ you see that it gives you the decimal representation of $x$ starting with the $n'th$ digit.
Example: let $x=\pi$, then we get $a(0)=0.141592..., a(1)=0.41592..., a(2)=0.1592...$ and so on.

Also this is a plot of $d(n)$ for $x=\pi$:

You see that for irrational x the height of the function at integers is not constant, nor periodic.

Thank you for your answer. Yes you are right but I think the fact that it is uniformly distributed is stronger than what I want to show. I hope that there is a simpler proof that can live without integrals :-)
–
ListingJan 2 '11 at 22:30

@user3123: No idea if a simpler proof exists. Where did you get this problem?
–
AryabhataJan 3 '11 at 6:15

Our Analysis Teacher gave us a holiday sheet we could make if we wanted and this has prooved to be the only tough question on it. I investigated a bit further and found this -> at.yorku.ca/cgi-bin/… on a forum about topology. I guess it uses math.stackexchange.com/questions/10280/… to show that there is a sequence that has a limit in every open interval. Then you could choose a interval of length $2^{-n}$ to prove it i think? But I don't fully understand it
–
ListingJan 3 '11 at 10:47

@user: That works, if a set A is dense in an interval, then for even real number r in that interval, there is a sequence of elements of A which converges to r. Remember the proof that for every real number there is a sequence of rationals converging to it?
–
AryabhataJan 3 '11 at 17:14